Question

доказать что n^3+3n^2+5n+3 нацело делиться на 3 методом математической индукции

Answer 1

1 шаг n=1 n^3+3n^2+5n+3=12 делится на 3.

2 шаг Пусть n=k k^3+3k^2+5k+3 делится на 3.

3 шаг n=k+1  (k+1)^3+3(k+1)^2+5(k+1)+3=

= k^3+3k^2+5k+3  +3k^2+3k+1+6k+3+5=(k^3+3k^2+5k+3) +3(k^2+2k+3) делится на 3, так как (k^3+3k^2+5k+3)делится на 3 по шагу 2 а 3(k^2+2k+3) делится на 3 из-за множетеля 3. ЧТД!

Answer 2

Proof that n^3 + 3n^2 + 5n + 3 is divisible by 3 using mathematical induction

To prove that the expression n^3 + 3n^2 + 5n + 3 is divisible by 3 for all positive integers n, we will use the method of mathematical induction.

Step 1: Base Case We start by verifying the statement for the base case, which is usually n = 1. Let's substitute n = 1 into the expression and check if it is divisible by 3:

1^3 + 3(1)^2 + 5(1) + 3 = 1 + 3 + 5 + 3 = 12

Since 12 is divisible by 3, the statement holds true for the base case.

Step 2: Inductive Hypothesis Assume that the statement is true for some positive integer k, i.e., k^3 + 3k^2 + 5k + 3 is divisible by 3.

Step 3: Inductive Step We need to prove that the statement holds for the next positive integer, which is k + 1. Let's substitute n = k + 1 into the expression and simplify:

(k + 1)^3 + 3(k + 1)^2 + 5(k + 1) + 3 = k^3 + 3k^2 + 3k + 1 + 3(k^2 + 2k + 1) + 5k + 5 + 3 = (k^3 + 3k^2 + 5k + 3) + (3k^2 + 6k + 4) = (k^3 + 3k^2 + 5k + 3) + 3(k^2 + 2k + 1 + 1)

From the inductive hypothesis, we know that k^3 + 3k^2 + 5k + 3 is divisible by 3. Additionally, 3(k^2 + 2k + 1 + 1) is also divisible by 3 since it is a multiple of 3. Therefore, the entire expression is divisible by 3.

Step 4: Conclusion By the principle of mathematical induction, we have shown that for any positive integer n, the expression n^3 + 3n^2 + 5n + 3 is divisible by 3.